How to describe the dynamics of a magnetic monopole charge in the external EM field using a Lagrangian in terms of the EM potentials?

Question

The equation of motion of a magnetic charge in the fixed external electromagnetic field $\mathbf{E},\mathbf{B}$ is $$ \frac{d}{dt}(\gamma m \mathbf{v})=q_m(\mathbf{B}-\mathbf{v}\times\mathbf{E}), $$ where $\gamma=\frac{1}{\sqrt{1-|\mathbf{v}|^2}}$ (using $c=\epsilon_0=\mu_0=1$), $q_m$ is the magnetic charge.

Of course, if one introduces the dual EM potential $(C^0,\mathbf{C})$, with $\mathbf{B}=-\nabla C^0-\dot{\mathbf{C}}$ and $\mathbf{E}=-\nabla\times\mathbf{C}$, then this equation of motion can be derived from the Lagrangian: $$ L=-\frac{m}{\gamma}-q_m(C^0-\mathbf{C}\cdot\mathbf{v}). $$ But is it possible to use instead a Lagrangian written in the original EM potential $(A^0,\mathbf{A})$, with $\mathbf{B}=\nabla\times\mathbf{A}$ and $\mathbf{E}=-\nabla A^0-\dot{\mathbf{A}}$ ? In other words, can we describe the dynamics of $\frac{d}{dt}(\gamma m \mathbf{v})=q_m(\mathbf{B}-\mathbf{v}\times\mathbf{E})+q_e(\mathbf{E}+\mathbf{v}\times\mathbf{B})$ with a Lagrangian?

Answer 1

You have a Lagrangian for electric charges, which is defined in terms of the potential $A$, ($\mu_0=1$) $$ \mathcal{L}_e = -\dfrac{1}{4}F_{\mu\nu}F^{\mu\nu} - j^\mu A_\mu, $$ where $F_{\mu\nu} := \partial_\mu A_\nu - \partial_\nu A_\mu$ and $j^\mu$ is the electric current. The equivalent one for magnetic charges, defined in terms of the potential $C$, $$ \mathcal{L}_m = -\dfrac{1}{4}F^*_{\mu\nu}F^{*\mu\nu} - m^\mu C_\mu, $$ where $m^\mu$ is the magnetic current, and $F^*_{\mu\nu} := \partial_\mu C_\nu - \partial_\nu C_\mu = \frac{1}{2}\epsilon_{\mu\nu\rho\sigma} F^{\rho\sigma}$ is the dual EM strength field. This gives us a relationship between $A$ and $C$: $$ \partial_\mu C_\nu - \partial_\nu C_\mu = \epsilon_{\mu\nu\rho\sigma} \partial^\rho A^\sigma $$

Now, the complete Lagrangian of the theory is $$ \mathcal{L}_t := \mathcal{L}_e - \mathcal{L}_m, $$ where we have importantly taken the difference to avoid a cancelation of the kinetic terms, as $F_{\mu\nu}F^{\mu\nu} = -F^*_{\mu\nu}F^{*\mu\nu}$. Thus the final Lagrangian is $$ \mathcal{L}_t = -\dfrac{1}{2}F_{\mu\nu}F^{\mu\nu} - j^\mu A_\mu + m^\mu C_\mu. $$

I don't know if you can do better than this.